In the circuit given below, the value of resistance X , when the...

$P = 15 + 6 = 21Ω, Q = \dfrac{8X}{8 + X} + 3$

$R = 15 + \dfrac{6 \times 6}{6 + 6} = 18Ω, S = 4 + \dfrac{4 \times 4}{4 + 4} = 6Ω$

Since potential difference across $BD$ is zero hence it satisfies Wheatstone's Bridge condition,

$∴ \dfrac{P}{Q} = \dfrac{R}{S},$ so $Q = \dfrac{PS}{R} = \dfrac{21 \times 6}{18} = 7Ω$

$3 + \dfrac{8x}{8 + x} = 7 ⇒ x = 8Ω$